# Reasonably Strong Algorithm
### Category: crypto
### Description:
[RSA](rsa.txt) strikes again!
### Author: lamchcl

### Solution:
Just another traditional rsa challenge, solved as usual by factorizing n with factordb and reconstructing the private key
```py
from Crypto.Util.number import inverse
from binascii import unhexlify

n = 126390312099294739294606157407778835887
e = 65537
c = 13612260682947644362892911986815626931

p = 9336949138571181619
q = 13536574980062068373

phi = (p-1)*(q-1)
d = inverse(e, phi)

m = pow(c, d, n)

print(unhexlify(hex(m)[2:]))
```
### Flag:
```
actf{10minutes}
```

Original writeup (https://github.com/r00tstici/writeups/blob/master/angstromCTF_2020/reasonably_strong_algorithm/README.md).